Move Last Element to Front of a Given Linked List

Move Last Element to Front of a Given Linked List

Problem statement: Given a Singly linked list, move the last element to the front of the linked list.

For example,

Here we traverse the linked list till the next pointer is NULL, store the last and second last element in two variables.

When we reach NULL pointer, assign the second last element’s next to NULL pointer.

Assign last element’s next to head.

At last assign the head to last element.

     head = last;

[code language=”cpp”]
/*C code to move the last element to front of the Linked List */
#include <stdio.h>
#include <stdlib.h>

typedef struct node
{
int data;
struct node *next;
} NODE;
NODE *head = NULL;

NODE *newNode(int key)
{
NODE *temp = (NODE *)malloc(sizeof(NODE));
temp->data = key;
temp->next = NULL;
return temp;
}

void moveToFront()
{

if (head == NULL || (head)->next == NULL)
return;
NODE *secLast = NULL;
NODE *last = head;

while (last->next != NULL)
{
secLast = last;
last = last->next;
}

secLast->next = NULL;
last->next = head;
head = last;
}

void printList()
{
NODE *temp = head;
if (temp == NULL)
{
printf("List is empty!\n");
return;
}
printf("Linked List is\t");
while (temp != NULL)
{
printf("%d–> ", temp->data);
temp = temp->next;
}
printf("NULL\n");
}

int main()
{
head = newNode(1);
head->next = newNode(2);
head->next->next = newNode(3);
head->next->next->next = newNode(4);
head->next->next->next->next = newNode(5);
head->next->next->next->next->next = NULL;
printList();
moveToFront();
printf("Linked list after moving last element to front\n");
printList();
return 0;
}
[/code]

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